I admit I had this problem on a set, but it's since been due and no one was able to solve it. I've not cracked it either, so let's see what you guys think.(adsbygoogle = window.adsbygoogle || []).push({});

F is a field, p is a prime. K is a Galois extension over F whose Galois group is a p-group (we call an extension of order p^n for some n a p-extension). L is a p-extension (not necessarily Galois) of K (and hence of F). Prove that the Galois closure of L is a p-extension of F.

It's very easy to construct a counter-example when K/F is not Galois. For example, let F be the rationals, let K be the splitting field of x^3-2, and let L be K adjoin a root of x^9-2. [K:F]=3, [L:K]=3, but the Galois closure has order 18 over F. So the theorem very much depends on the Galois-ity of K/F.

Intuitively, I feel like if (letting L' be the Galois closure of L) |Gal(L'/F)|=[L':F] has a divisor that is not a p-power, since |Gal(L'/K)| must be simple, we can use Sylow's theorem to show a contradiction. Other thoughts include that if we can find Galois extensions of K that give irreducible factors for the minimal polynomial of a (where L=K(a) by the primitive element theorem), then these factors all have to have the same degree (by considering the action of a normal subgroup on a set that the parent group acts transitively on), hence they have to have p-power degree. Some kind of induction would show the minimal polynomial has a p-power splitting field.

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# Galois Closure of p-Extensions

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